Stuck on Thevenin Equivalence Problem [closed]

Question

I'm supposed to find the necessary elements for the Thevenin equivalent of the pictured circuit. That's \$V_{TH}\$, \$R_{TH}\$, and the Norton current (\$I_N\$). I'm also required to show that \$\frac{V_{TH}}{R_{TH}} = I_N\$. This is where I run into problems. From what I've been able to find, \$V_{TH} = I R_2\$, \$R_{TH} = R_1\$, and \$I_N = \frac{R_2}{R_1 + R_2}\$. This clearly doesn't obey the equation above. However, I haven't been able to find any other results for each variable, and they all seem to match up with simulations of the circuit. I've been stuck here for a while, please let me know if I'm missing something.

Answer 1

Your \$R_{TH}\$ seems to be incorrect. When we remove the current source (replacing it by an open circuit), we essentially get between \$A\$ and \$B\$ two resistors in series. In particular, we have \$R_{TH} = R_1 \color{red}{+ R_2}\$.

Then if we want to find \$I_N\$ independently, we find the short-circuit current for terminal \$A - B\$. The circuit is just a current divider. We get \$I_N = \frac{R_2}{R_1 + R_2} \color{red}{I}\$.

Now, we can verify that

$$\frac{V_{TH}}{R_{TH}} = \frac{I R_2}{R_1 + R_2} = \frac{R_2}{R_1 + R_2} I = I_N$$

as desired.

Answer 2

To find the thevenin equivalent resistance, open circuit the current source.short circuit any voltage sources while keeping dependent current and voltage sources as they are. In the circuit above, the thevenin resistance is \$R_{Th} = R_1 + R_2\$.

The Thevenin voltage across a-b is \$IR_2\$.

You can find \$\frac{V_{Th}}{R_{Th}}\$ from above.

Answer 3

addressing your question

From what I've been able to find, \$V_{_\text{TH}}=I\cdot R_2\$, \$R_{_\text{TH}}=R_1\$

Your \$I\$ and \$R_2\$ is already a Norton source. The Norton source can be converted to its equivalent Thevenin source. Note that \$R_1\$ isn't needed to do that. This is entirely about \$I\$ and \$R_2\$:

^{simulate this circuit – Schematic created using CircuitLab}

You are correct that \$V_{_\text{TH}}=I\cdot R_2\$. But incorrect when writing that \$R_{_\text{TH}}=R_1\$. Instead, \$R_{_\text{TH}}=R_2\$.

If you look, you should see that \$R_1\$ wasn't part of the conversion.

The goal is to have a single ideal voltage source and a single ideal resistance. So you need to add \$R_1\$ to \$R_{_\text{TH}}\left[=R_2\right]\$ together to get the final Thevenin equivalent resistance.

Let's call this value \$R_{_\text{TH}}^{\:'}=R_{_\text{TH}}+R_1=R_2+R_1\$.

You can convert this back to a Norton equivalent. The new ideal Norton current source will be \$I_{_\text{N}}=\frac{V_{_\text{TH}}}{R_{_\text{TH}}^{\:'}}\$. The Norton resistance will just be \$R_{_\text{N}}=R_{_\text{TH}}^{\:'}\$.